Probably yes, this problem has connection to sphere packing. But how to approach this problem?
I can calculate length of base side of a tetrahedron and divide it by sphere's diameter. From that I know how many spheres will be in the base level and then I can calculate next levels. But is this a good approach?

Imagine 3 identical spheres of diameter d, all tangent and whose centers lie in the same horizontal plane. Now we place a 4th sphere whose diameter is also d on top so that it is tangent to the other three. If we draw line segments joining the centers of the 4 spheres we form a regular tetrahedron whose side lengths are d. Now, to find the height h of the tetrahedron, we may drop a vertical line segment from the apex to the base and label its length h. Next draw a line segment from the apex down one of its faces bisecting the face, and we find its length from . Now, observing that the vertical line h intersects the base in the middle (the point equidistant from the 3 vertices), we can find the length of the line segment b joining h and l from dividing the equilateral base into 3 congruent isosceles triangles each of which has an area 1/3 that of the equilateral base:

Now we have a right triangle whose legs are h and b and whose hypotenuse is l, thus from the Pythagorean theorem, we obtain:

From this we may determine how many levels n:

We are given and we wish to form a tetrahedral stack high. Hence:

Our stack will be just over 6 ft. high.

Thus, the total number of balls S is found by adding up the first 10 triangular numbers: